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Design and Simulation of Two Stroke Engines

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<strong>Design</strong> <strong>and</strong> <strong>Simulation</strong> <strong>of</strong> <strong>Two</strong>-<strong>Stroke</strong> <strong>Engines</strong><br />

From the previous discussion regarding the relevance <strong>of</strong> specific time area, the numerical<br />

level for the intake in question as a function <strong>of</strong> bmep is given by Eq. 6.1.9 <strong>and</strong> repeated here<br />

as:<br />

bmep + 1.528<br />

A svi = — s / m (6.4.5)<br />

When the target requirement <strong>of</strong> bmep for a given swept volume, Vsv, at an engine speed,<br />

rpm, is declared, there is available a direct solution for the design <strong>of</strong> the entire device. From<br />

Eq. 6.1.12:<br />

2>ede<br />

Specific time area, Asvi = ^ s s/m (6.4.6)<br />

6Vsvrpm<br />

If you glance ahead at Fig. 6.29, the data listed in the upper left-h<strong>and</strong> side <strong>of</strong> that figure<br />

contain all <strong>of</strong> the relevant input data for the calculation, as exhibited in Eqs. 6.4.1 to 6.4.6. It<br />

is seen that the unknown data value which emerges from the calculation is the value <strong>of</strong> the<br />

outer port edge radius, rmax. This is the most convenient method <strong>of</strong> solution. Eqs. 6.4.1-6.4.6<br />

are combined, estimates are made <strong>of</strong> all data but rmax, <strong>and</strong> the result is a direct solution for<br />

rmax. This is not as difficult as it might appear, for disc valve timings are not subject to great<br />

variations from the lowest to the highest specific power output level. For example, the timing<br />

events, 0dVo an( ^ 9dvc> would be at 150 <strong>and</strong> 65 for a small scooter engine <strong>and</strong> at 140 <strong>and</strong> 80,<br />

respectively, for a racing engine. For completeness, the actual solution route for the determination<br />

<strong>of</strong> rmax <strong>and</strong> the carbureter flow diameter (from Fig. 5.5), dtv, is given below:<br />

1 Ad6 = 6AsviVsvrpm (6.4.7)<br />

fAdO<br />

A max = ~ — (6.4.8)<br />

b max ~ ^p<br />

360 CcAmax + rp 2 (4 - it))<br />

r max = J " + r min (6.4.9)<br />

d = 4C c A max (64 10)<br />

V n<br />

458

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